# Flamant solution

Flamant solution

The Flamant solution provides expressions for the stresses and displacements in a linear elastic wedge loaded by point forces at its sharp end. This solution was developed by A. Flamant [A. Flamant. (1892). "Sur la répartition des pressions dans un solide rectangulaire chargé transversalement." Compte. Rendu. Acad. Sci. Paris, vol. 114, p. 1465.] in 1892 by modifying the three-dimensional solution of Boussinesq.

The stresses predicted by the Flamant solution are (in polar coordinates):where $C_1, C_3$ are constants that are determined from the boundary conditions and the geometry of the wedge (i.e., the angles ) and satisfy:where $F_1,F_2$ are the applied forces. The wedge problem is "self-similar" and has no inherent length scale. Also, all quantities can be expressed in the separated-variable form $sigma = f\left(r\right)g\left( heta\right)$. The stresses vary as $\left(1/r\right)$.

Forces acting on a half-plane

For the special case where $alpha = -pi$, , the wedge is converted into a half-plane with a normal force and a tangential force. In that case:$C_1 = - frac\left\{F_1\right\}\left\{pi\right\} ~;~~ C_3 = -frac\left\{F_2\right\}\left\{pi\right\}$Therefore the stresses are:and the displacements are (using Michell's solution):The $ln r$ dependence of the displacements implies that the displacement grows the further one moves from the point of application of the force (and is unbounded at infinity). This feature of the Flamant solution is confusing and appears unphysical. For a discussion of the issue see [http://imechanica.org/node/319 http://imechanica.org/node/319] .

Displacements at the surface of the half-plane

The displacements in the $x_1, x_2$ directions at the surface of the half-plane are given by:where:$u$ is the Poisson's ratio, $mu$ is the shear modulus, and:

Derivation of Flamant solution

If we assume the stresses to vary as $\left(1/r\right)$, we can pick terms containing $1/r$ in the stresses from Michell's solution. Then the Airy stress function can be expressed as:$varphi = C_1 r hetasin heta + C_2 rln r cos heta + C_3 r hetacos heta + C_4 rln r sin heta$Therefore, from the tables in Michell's solution, we have:The constants $C_1, C_2, C_3, C_4$ can then, in principle, be determined from the wedge geometry and the applied boundary conditions.

However, the concentrated loads at the vertex are difficult to express in terms of traction boundary conditions because
# the unit outward normal at the vertex is undefined
# the forces are applied at a point (which has zero area) and hence the traction at that point is infinite.

To get around this problem, we consider a bounded region of the wedge and consider equilibrium of the bounded wedge. [Slaughter, W. S. (2002). "The Linearized Theory of Elasticity". Birkhauser, Boston, p. 294.] [ J. R. Barber, 2002, "Elasticity: 2nd Edition", Kluwer Academic Publishers. ] Let the bounded wedge have two traction free surfaces and a third surface in the form of an arc of a circle with radius $a,$. Along the arc of the circle, the unit outward normal is $mathbf\left\{n\right\} = mathbf\left\{e\right\}_r$ where the basis vectors are $\left(mathbf\left\{e\right\}_r, mathbf\left\{e\right\}_ heta\right)$. The tractions on the arc are:

Next, we examine the force and moment equilibrium in the bounded wedge and get: We require that these equations be satisfied for all values of $a,$ and thereby satisfy the boundary conditions.

The traction-free boundary conditions on the edges $heta=alpha$ and also imply that:except at the point $r = 0$.

If we assume that $sigma_\left\{r heta\right\}=0$ everywhere, then the traction-free conditions and the moment equilibrium equation are satisfied and we are left with :and $sigma_\left\{ heta heta\right\} = 0$ along except at the point $r = 0$. But the field $sigma_\left\{ heta heta\right\} = 0$ everywhere also satisfies the force equilibrium equations. Hence this must be the solution. Also, the assumption $sigma_\left\{r heta\right\}=0$ implies that $C_2 = C_4 = 0$.

Therefore,:$sigma_\left\{rr\right\} = frac\left\{2C_1cos heta\right\}\left\{r\right\} + frac\left\{2C_3sin heta\right\}\left\{r\right\} ~;~~sigma_\left\{r heta\right\} = 0 ~;~~ sigma_\left\{ heta heta\right\} = 0$

To find a particular solution for $sigma_\left\{rr\right\}$ we have to plug in the expression for $sigma_\left\{rr\right\}$ into the force equilibrium equations to get a system of two equations which have to be solved for $C_1, C_3$::

Forces acting on a half-plane

If we take $alpha = -pi$ and , the problem is converted into one where a normal force $F_2$ and a tangential force $F_1$ act on a half-plane. In that case, the force equilibrium equations take the form:Therefore:$C_1 = - cfrac\left\{F_1\right\}\left\{pi\right\} ~;~~ C_3 = - cfrac\left\{F_2\right\}\left\{pi\right\} ~.$The stresses for this situation are:$sigma_\left\{rr\right\} = -frac\left\{2\right\}\left\{pi r\right\}\left(F_1cos heta + F_2sin heta\right) ~;~~sigma_\left\{r heta\right\} = 0 ~;~~ sigma_\left\{ heta heta\right\} = 0$Using the displacement tables from the Michell solution, the displacements for this case are given by:

Displacements at the surface of the half-plane

To find expressions for the displacements at the surface of the half plane, we first find the displacements for positive $x_1$ ($heta=0$) and negative $x_1$ ($heta = pi$) keeping in mind that $r = |x_1|$ along these locations.

For $heta=0$ we have:For $heta = pi$ we have:We can make the displacements symmetric around the point of application of the force by adding rigid body displacements (which does not affect the stresses) :$u_1 = cfrac\left\{F_2\right\}\left\{8mu\right\}\left(kappa-1\right) ~;~~ u_2 = cfrac\left\{F_1\right\}\left\{8mu\right\}\left(kappa-1\right)$and removing the redundant rigid body displacements:$u_1 = cfrac\left\{F_1\right\}\left\{4pimu\right\} ~;~~ u_2 = cfrac\left\{F_2\right\}\left\{4pimu\right\} ~.$Then the displacements at the surface can be combined and take the form:where:

References

* Michell solution
* Linear elasticity
* Stress (physics)

Wikimedia Foundation. 2010.

### Look at other dictionaries:

• Flamant — is the French for flamingoIt may refer to *The Dassault MD 315 Flamant, an aircraft *Flamant class patrol vessel, a type of ship *Flamant solution, the solution to a problem in linear elasticity provided by A. Flamant in 1892 …   Wikipedia

• Michell solution — The Michell solution is a general solution to the elasticity equations in polar coordinates ( ). The solution is such that the stress components are in the form of a Fourier series in . Michell[1] showed that the general solution can be expressed …   Wikipedia

• Contact mechanics — Continuum mechanics …   Wikipedia

• Expert system — In artificial intelligence, an expert system is a computer system that emulates the decision making ability of a human expert.[1] Expert systems are designed to solve complex problems by reasoning about knowledge, like an expert, and not by… …   Wikipedia

• Alice au pays des merveilles (film, 1951) — Pour les articles homonymes, voir Alice au pays des merveilles (homonymie). Alice au pays des merveilles Données clés Titre original Alice in Wonderland Réalisation Clyde Geronimi Wilfred Jackson …   Wikipédia en Français

• Dassault LOGIDUC — The Dassault LOGIDUC sometimes spelled Logiduc in French and LogiDuc in English (Logique de Développement d UCAV, French for Unmanned Combat Aerial Vehicle development solution ) was an autonomous industrial program launched in 1999 by the French …   Wikipedia

• Dassault Mirage IV — …   Wikipédia en Français

• Joseph Boussinesq — Naissance 13 mars 1842 Saint André de Sangonis (France) Décès 19 février 1929 Paris (France) Nationalité …   Wikipédia en Français

• Neoplatonisme — Néoplatonisme ██████████ …   Wikipédia en Français

• Néo-Platonisme — Néoplatonisme ██████████ …   Wikipédia en Français